U s i n g P = Work Time Using\ P = \frac{\text{~Work~}}{\text{~Time~}} P 1 = m g h t 1 = ( 300 ) g ( 100 ) 5 P_{1} = \frac{mgh}{t_{1}} = \frac{(300)g(100)}{5} P 2 = ( 50 ) g ( 100 ) 2 P_{2} = \frac{(50)g(100)}{2} P 1 P 2 = 600 250 = 12 5 = 3 × 4 4 + 1 = 3 16 16 + 1 \frac{P_{1}}{P_{2}} = \frac{600}{250} = \frac{12}{5} = \frac{3 \times 4}{4 + 1} = \frac{3\sqrt{16}}{\sqrt{16} + 1} T h u s x = 16 Thus\ x = 16

The ratio of powers of two motors is 3√x/(√x+1), that are capable of raising 300 kg water in 5 minutes and 50 kg water in 2 minutes respectively from a well of 100 m deep. The value of x will be

Question 1:

The ratio of powers of two motors is 3√x/(√x+1), that are capable of raising 300 kg water in 5 minutes and 50 kg water in 2 minutes respectively from a well of 100 m deep. The value of x will be

2.4

Answer:

Solution:

$Using\ P = \frac{\text{~Work~}}{\text{~Time~}}$

$P_{1} = \frac{mgh}{t_{1}} = \frac{(300)g(100)}{5}$

$P_{2} = \frac{(50)g(100)}{2}$

$\frac{P_{1}}{P_{2}} = \frac{600}{250} = \frac{12}{5} = \frac{3 \times 4}{4 + 1} = \frac{3\sqrt{16}}{\sqrt{16} + 1}$

$Thus\ x = 16$

Prev Exam(s):

JEE MAIN

- 2023

, 13 Apr Shift-01

Chapter:

WORK, POWER AND ENERGY (WPE)

Question Level:

Tags:

PHYSICS
WORK, POWER AND ENERGY (WPE)

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